Comparing Optical Signal Power with RMS Voltage

I have a CW laser system, the optical detector receives a pulse width modulated signal and converts it to an analog signal. Both the source and detector have 32 channels and I would like to know what the analog signal power is, I thought maybe I could use the RMS voltage received on the oscilloscope and convert that to dB. I can then compare the input and output signals, is this possible?

In most applications, you wish to understand what your signal to noise ratio is which is simple if your modulation is simple. Mulitplexed and pulsed modulation schemes become very difficult as the RMS voltage is dependent upon your modulation.

You may wish to look at the RMS background noise and multiply it by 6.6 to convert that to a peak-peak background noise.
Then, measure your peak-to-peak from modulation.

SNR (dB) = 20 * log (peak-peak_mod / peak-peak_noise)

For a part to do monitor your output strength, you may wish to look into the LTC5507 low frequency peak power detector.

Fun project. I used to mux signals in high school, and acquired a fiber optic link in college to send stereo using PWM :)

I have a CW laser system, the optical detector receives a pulse width modulated signal and converts it to an analog signal. Both the source and detector have 32 channels and I would like to know what the analog signal power is, I thought maybe I could use the RMS voltage received on the oscilloscope and convert that to dB. I can then compare the input and output signals, is this possible?

I think you mean the signal power of one analogue output. If the modulation is sinusoidal, you can find the RMS voltage using an oscilloscope. You must have the correct load resistor in place. Then the power in the load (Pr) is Vrms^2 / R. If wanted, you can then measure the analogue signal power into the TX (Pt) in a similar way (you need to know the input resistance). If you want overall loss in decibels it is 10 log (Pt/Pr).

This type of signal is an asymmetric pulse rather than sine wave. Essentially, you cannot have a "negative" amount of light. Peak-Peak measurements are more applicable.
However, if your sensor's output reflects cable power, then the 10 log (Vout / Vnoise) reflects the signal to noise in terms of cable power.
If your addressing the noise power in terms of output voltage, then the 20 log(Vout / Vnoise).
Tech99's suggestion holds valid for computing the effects of fiber changes (i.e. more cable), so I suspect it is most valid.
Mike

Thanks everyone for your help, I was able to reproduce the results while testing my systems which allowed me to easily determine the quality of the analog output signal. One issue that I ran into was that some of the channels on the laser detectors were sending out inverse signals but when looking at the RMS on the oscilloscope based off my calculations the signal should of been correct. I decided to take the ratio of the RMS and the amplitude which allowed me to correct this issue.